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Question: How to solve the equation \[{{z}^{4}}=i\overline{z}\] , where z is a complex number?...

How to solve the equation z4=iz‾{{z}^{4}}=i\overline{z} , where z is a complex number?

Explanation

Solution

To solve the equation z4=iz‾{{z}^{4}}=i\overline{z} , we have to assume z=reiθz=r{{e}^{i\theta }} . Then, z‾\overline{z} is the complex conjugate, that is, z‾=re−iθ\overline{z}=r{{e}^{-i\theta }} . Now, we have to substitute these values in the given equation as simplify. We will change the exponent because complex exponents have a period of 2π2\pi . So, we will get (reiθ)4=i(rei(−θ+2nπ)),n∈N{{\left( r{{e}^{i\theta }} \right)}^{4}}=i\left( r{{e}^{i\left( -\theta +2n\pi \right)}} \right),n\in \mathbb{N} . Now, we will further simplify this equation and equate the modulus and argument parts (or real and imaginary parts). Then, we will find the value of r and corresponding values of z for various values of n.

Complete step-by-step solution:
We have to solve the equation z4=iz‾{{z}^{4}}=i\overline{z} . We are given that z is a complex number. Let us assume z=reiθz=r{{e}^{i\theta }} . Then, z‾\overline{z} is the complex conjugate, that is, z‾=re−iθ\overline{z}=r{{e}^{-i\theta }} . Now, let us substitute these in the given equation.
⇒(reiθ)4=i(re−iθ)\Rightarrow {{\left( r{{e}^{i\theta }} \right)}^{4}}=i\left( r{{e}^{-i\theta }} \right)
We know that complex exponents have has a period of 2π2\pi . Therefore, we can write the above equation as
⇒(reiθ)4=i(rei(−θ+2nπ)),n∈N\Rightarrow {{\left( r{{e}^{i\theta }} \right)}^{4}}=i\left( r{{e}^{i\left( -\theta +2n\pi \right)}} \right),n\in \mathbb{N}
We know that eiπ2=cos⁡π2+isin⁡π2=i{{e}^{i\dfrac{\pi }{2}}}=\cos \dfrac{\pi }{2}+i\sin \dfrac{\pi }{2}=i .
⇒(reiθ)4=eiπ2(rei(−θ+2nπ))\Rightarrow {{\left( r{{e}^{i\theta }} \right)}^{4}}={{e}^{i\dfrac{\pi }{2}}}\left( r{{e}^{i\left( -\theta +2n\pi \right)}} \right)
We know that (ab)m=am×bm{{\left( ab \right)}^{m}}={{a}^{m}}\times {{b}^{m}} and (am)n=amn{{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}} .
⇒r4ei4θ=eiπ2(rei(−θ+2nπ))\Rightarrow {{r}^{4}}{{e}^{i4\theta }}={{e}^{i\dfrac{\pi }{2}}}\left( r{{e}^{i\left( -\theta +2n\pi \right)}} \right)
We know that am×an=am+n{{a}^{m}}\times {{a}^{n}}={{a}^{m+n}} .
⇒r4ei4θ=rei(π2−θ+2nπ)\Rightarrow {{r}^{4}}{{e}^{i4\theta }}=r{{e}^{i\left( \dfrac{\pi }{2}-\theta +2n\pi \right)}}
Now, let us equate the modulus and argument parts (or real and imaginary parts). When we equate the real parts, we will get
r4=r...(i){{r}^{4}}=r...\left( i \right)
When we equate the imaginary, we will get
4θ=π2−θ+2nπ...(ii)4\theta =\dfrac{\pi }{2}-\theta +2n\pi ...\left( ii \right)
From equation (i), we can find the value of r. Let us take r from the RHS to the LHS.
⇒r4−r=0\Rightarrow {{r}^{4}}-r=0
Let us take common r outside.
⇒r(r3−1)=0\Rightarrow r\left( {{r}^{3}}-1 \right)=0
This means, we can write either r=0r=0 or r3−1=0{{r}^{3}}-1=0 . Let us consider r3−1=0{{r}^{3}}-1=0 . We have to take -1 to the RHS.
⇒r3=1\Rightarrow {{r}^{3}}=1
Let us take the cube root of the above equation.
⇒r=1\Rightarrow r=1
Therefore, we obtained r=0,1r=0,1 .
So, when r=0r=0 , we can write
z=reiθ=0×eiθ ⇒z=0 \begin{aligned} & z=r{{e}^{i\theta }}=0\times {{e}^{i\theta }} \\\ & \Rightarrow z=0 \\\ \end{aligned}
When r=1r=1 , we will get
z=reiθ=1×eiθ ⇒z=eiθ \begin{aligned} & z=r{{e}^{i\theta }}=1\times {{e}^{i\theta }} \\\ & \Rightarrow z={{e}^{i\theta }} \\\ \end{aligned}
Now, let us simplify the equation (ii) by taking θ\theta to the LHS.
⇒4θ+θ=π2+2nπ ⇒5θ=π2+2nπ \begin{aligned} & \Rightarrow 4\theta +\theta =\dfrac{\pi }{2}+2n\pi \\\ & \Rightarrow 5\theta =\dfrac{\pi }{2}+2n\pi \\\ \end{aligned}
Let us simplify the RHS.
⇒5θ=π2+4nπ2 ⇒5θ=π2(1+4n) \begin{aligned} & \Rightarrow 5\theta =\dfrac{\pi }{2}+\dfrac{4n\pi }{2} \\\ & \Rightarrow 5\theta =\dfrac{\pi }{2}\left( 1+4n \right) \\\ \end{aligned}
We have to take 5 to the RHS.
⇒θ=π10(1+4n)\Rightarrow \theta =\dfrac{\pi }{10}\left( 1+4n \right)
Let us substitute various values of n, that is, 0, 1, 2, … in the above equation and find the corresponding values of z when r=1r=1, that is, z=eiθz={{e}^{i\theta }} .
We know that eiθ=cos⁡θ+isin⁡θ{{e}^{i\theta }}=\cos \theta +i\sin \theta .
⇒z=cos⁡θ+isin⁡θ\Rightarrow z=\cos \theta +i\sin \theta
When n=0n=0 , we will get
θ=π10\theta =\dfrac{\pi }{10}
∴z=cos⁡π10+isin⁡π10\therefore z=\cos \dfrac{\pi }{10}+i\sin \dfrac{\pi }{10}
We know that
⇒cos⁡x=2cos⁡2x2−1 ⇒cos⁡2x2=cos⁡x+12 ⇒cos⁡x2=cos⁡x+12 \begin{aligned} & \Rightarrow \cos x=2{{\cos }^{2}}\dfrac{x}{2}-1 \\\ & \Rightarrow {{\cos }^{2}}\dfrac{x}{2}=\dfrac{\cos x+1}{2} \\\ & \Rightarrow \cos \dfrac{x}{2}=\sqrt{\dfrac{\cos x+1}{2}} \\\ \end{aligned}
Now, we can write cos⁡π10=cos⁡(π5)2=cos⁡π5+12...(iii)\cos \dfrac{\pi }{10}=\cos \dfrac{\left( \dfrac{\pi }{5} \right)}{2}=\sqrt{\dfrac{\cos \dfrac{\pi }{5}+1}{2}}...\left( iii \right)

We know that
2π5=π−3π5 ⇒sin⁡(2π5)=sin⁡(π−3π5) ⇒sin⁡(2π5)=sin⁡(3π5) \begin{aligned} & \dfrac{2\pi }{5}=\pi -\dfrac{3\pi }{5} \\\ & \Rightarrow \sin \left( \dfrac{2\pi }{5} \right)=\sin \left( \pi -\dfrac{3\pi }{5} \right) \\\ & \Rightarrow \sin \left( \dfrac{2\pi }{5} \right)=\sin \left( \dfrac{3\pi }{5} \right) \\\ \end{aligned}
We know that sin⁡2θ=2sin⁡θcos⁡θ\sin 2\theta =2\sin \theta \cos \theta .
⇒2sin⁡(π5)cos⁡(π5)=3sin⁡(π5)−4sin⁡3(π5) ⇒2sin⁡(π5)cos⁡(π5)=sin⁡(π5)(3−4sin⁡2(π5)) ⇒2\requirecancelsin⁡(π5)cos⁡(π5)=\requirecancelsin⁡(π5)(3−4sin⁡2(π5)) ⇒2cos⁡(π5)=3−4sin⁡2(π5) \begin{aligned} & \Rightarrow 2\sin \left( \dfrac{\pi }{5} \right)\cos \left( \dfrac{\pi }{5} \right)=3\sin \left( \dfrac{\pi }{5} \right)-4{{\sin }^{3}}\left( \dfrac{\pi }{5} \right) \\\ & \Rightarrow 2\sin \left( \dfrac{\pi }{5} \right)\cos \left( \dfrac{\pi }{5} \right)=\sin \left( \dfrac{\pi }{5} \right)\left( 3-4{{\sin }^{2}}\left( \dfrac{\pi }{5} \right) \right) \\\ & \Rightarrow 2\require{cancel}\cancel{\sin \left( \dfrac{\pi }{5} \right)}\cos \left( \dfrac{\pi }{5} \right)=\require{cancel}\cancel{\sin \left( \dfrac{\pi }{5} \right)}\left( 3-4{{\sin }^{2}}\left( \dfrac{\pi }{5} \right) \right) \\\ & \Rightarrow 2\cos \left( \dfrac{\pi }{5} \right)=3-4{{\sin }^{2}}\left( \dfrac{\pi }{5} \right) \\\ \end{aligned}
We know that cos⁡2x=1−sin⁡2x{{\cos }^{2}}x=1-{{\sin }^{2}}x .
⇒2cos⁡(π5)=3−4(1−cos⁡2(π5))\Rightarrow 2\cos \left( \dfrac{\pi }{5} \right)=3-4\left( 1-{{\cos }^{2}}\left( \dfrac{\pi }{5} \right) \right)
Let us consider X=cos⁡(π5)X=\cos \left( \dfrac{\pi }{5} \right) .
⇒2X=3−4(1−X2) ⇒2X=3−4+4X2 ⇒2X=−1+4X2 ⇒4X2−2X−1 ⇒4X2−2X−1 \begin{aligned} & \Rightarrow 2X=3-4\left( 1-{{X}^{2}} \right) \\\ & \Rightarrow 2X=3-4+4{{X}^{2}} \\\ & \Rightarrow 2X=-1+4{{X}^{2}} \\\ & \Rightarrow 4{{X}^{2}}-2X-1 \\\ & \Rightarrow 4{{X}^{2}}-2X-1 \\\ \end{aligned}
Let us substitute 4X2−2X+1=04{{X}^{2}}-2X+1=0 . Then using quadratic formula, we can find the value of X.
X=−(−2)±(−2)2−4×4×−12×4 ⇒X=2±4+168 ⇒X=2±208 ⇒X=2±258 ⇒X=1±54 ⇒X=1+54,X=1−54 \begin{aligned} & X=\dfrac{-\left( -2 \right)\pm \sqrt{{{\left( -2 \right)}^{2}}-4\times 4\times -1}}{2\times 4} \\\ & \Rightarrow X=\dfrac{2\pm \sqrt{4+16}}{8} \\\ & \Rightarrow X=\dfrac{2\pm \sqrt{20}}{8} \\\ & \Rightarrow X=\dfrac{2\pm 2\sqrt{5}}{8} \\\ & \Rightarrow X=\dfrac{1\pm \sqrt{5}}{4} \\\ & \Rightarrow X=\dfrac{1+\sqrt{5}}{4},X=\dfrac{1-\sqrt{5}}{4} \\\ \end{aligned}
We know that cos⁡(π5)\cos \left( \dfrac{\pi }{5} \right) is positive.
⇒X=1+54 ⇒cos⁡(π5)=1+54 \begin{aligned} & \Rightarrow X=\dfrac{1+\sqrt{5}}{4} \\\ & \Rightarrow \cos \left( \dfrac{\pi }{5} \right)=\dfrac{1+\sqrt{5}}{4} \\\ \end{aligned}
Let us substitute this in (iii).
cos⁡π10==1+54+12 ⇒cos⁡π10=1+5+48 ⇒cos⁡π10=5+58 ⇒cos⁡π10=5+522 ⇒cos⁡π10=1212(5+5) \begin{aligned} & \cos \dfrac{\pi }{10}==\sqrt{\dfrac{\dfrac{1+\sqrt{5}}{4}+1}{2}} \\\ & \Rightarrow \cos \dfrac{\pi }{10}=\sqrt{\dfrac{1+\sqrt{5}+4}{8}} \\\ & \Rightarrow \cos \dfrac{\pi }{10}=\sqrt{\dfrac{5+\sqrt{5}}{8}} \\\ & \Rightarrow \cos \dfrac{\pi }{10}=\dfrac{\sqrt{5+\sqrt{5}}}{2\sqrt{2}} \\\ & \Rightarrow \cos \dfrac{\pi }{10}=\dfrac{1}{2}\sqrt{\dfrac{1}{2}\left( 5+\sqrt{5} \right)} \\\ \end{aligned}
Now, let us find sin⁡π10\sin \dfrac{\pi }{10} .
sin⁡π10=sin⁡(π5)2\sin \dfrac{\pi }{10}=\sin \dfrac{\left( \dfrac{\pi }{5} \right)}{2}
We know that
sin⁡2x=2sin⁡xcos⁡x ⇒sin⁡x=2sin⁡x2cos⁡x2 ⇒sin⁡x2=sin⁡x2cos⁡x2 \begin{aligned} & \sin 2x=2\sin x\cos x \\\ & \Rightarrow \sin x=2\sin \dfrac{x}{2}\cos \dfrac{x}{2} \\\ & \Rightarrow \sin \dfrac{x}{2}=\dfrac{\sin x}{2\cos \dfrac{x}{2}} \\\ \end{aligned}
Therefore, we can write
sin⁡(π5)2=sin⁡π52cos⁡π10\sin \dfrac{\left( \dfrac{\pi }{5} \right)}{2}=\dfrac{\sin \dfrac{\pi }{5}}{2\cos \dfrac{\pi }{10}}
We obtained cos⁡(π5)=1+54\cos \left( \dfrac{\pi }{5} \right)=\dfrac{1+\sqrt{5}}{4} .
We know that sin⁡2x=1−cos⁡2x{{\sin }^{2}}x=1-{{\cos }^{2}}x .
⇒sin⁡2(π5)=1−(1+54)2 ⇒sin⁡2(π5)=1−(1−25+(5)216) ⇒sin⁡2(π5)=1−(6−2516) ⇒sin⁡2(π5)=1−(3−58) ⇒sin⁡2(π5)=(8−3+58) ⇒sin⁡2(π5)=5−58 ⇒sin⁡(π5)=5−58 \begin{aligned} & \Rightarrow {{\sin }^{2}}\left( \dfrac{\pi }{5} \right)=1-{{\left( \dfrac{1+\sqrt{5}}{4} \right)}^{2}} \\\ & \Rightarrow {{\sin }^{2}}\left( \dfrac{\pi }{5} \right)=1-\left( \dfrac{1-2\sqrt{5}+{{\left( \sqrt{5} \right)}^{2}}}{16} \right) \\\ & \Rightarrow {{\sin }^{2}}\left( \dfrac{\pi }{5} \right)=1-\left( \dfrac{6-2\sqrt{5}}{16} \right) \\\ & \Rightarrow {{\sin }^{2}}\left( \dfrac{\pi }{5} \right)=1-\left( \dfrac{3-\sqrt{5}}{8} \right) \\\ & \Rightarrow {{\sin }^{2}}\left( \dfrac{\pi }{5} \right)=\left( \dfrac{8-3+\sqrt{5}}{8} \right) \\\ & \Rightarrow {{\sin }^{2}}\left( \dfrac{\pi }{5} \right)=\dfrac{5-\sqrt{5}}{8} \\\ & \Rightarrow \sin \left( \dfrac{\pi }{5} \right)=\sqrt{\dfrac{5-\sqrt{5}}{8}} \\\ \end{aligned}
Therefore, we can write
⇒sin⁡(π5)2=5−582cos⁡π10 ⇒sin⁡(π5)2=5−5825+58 ⇒sin⁡π10=5−582×1212(5+5) ⇒sin⁡π10=5−54(5+5) \begin{aligned} & \Rightarrow \sin \dfrac{\left( \dfrac{\pi }{5} \right)}{2}=\dfrac{\sqrt{\dfrac{5-\sqrt{5}}{8}}}{2\cos \dfrac{\pi }{10}} \\\ & \Rightarrow \sin \dfrac{\left( \dfrac{\pi }{5} \right)}{2}=\dfrac{\sqrt{\dfrac{5-\sqrt{5}}{8}}}{2\sqrt{\dfrac{5+\sqrt{5}}{8}}} \\\ & \Rightarrow \sin \dfrac{\pi }{10}=\dfrac{\sqrt{\dfrac{5-\sqrt{5}}{8}}}{2\times \dfrac{1}{2}\sqrt{\dfrac{1}{2}\left( 5+\sqrt{5} \right)}} \\\ & \Rightarrow \sin \dfrac{\pi }{10}=\sqrt{\dfrac{5-\sqrt{5}}{4\left( 5+\sqrt{5} \right)}} \\\ \end{aligned}
Let us take the conjugate.
⇒sin⁡π10=12(5−5)(5−5)(5+5)(5−5) ⇒sin⁡π10=12(5−5)2(5+5)(5−5) \begin{aligned} & \Rightarrow \sin \dfrac{\pi }{10}=\dfrac{1}{2}\sqrt{\dfrac{\left( 5-\sqrt{5} \right)\left( 5-\sqrt{5} \right)}{\left( 5+\sqrt{5} \right)\left( 5-\sqrt{5} \right)}} \\\ & \Rightarrow \sin \dfrac{\pi }{10}=\dfrac{1}{2}\sqrt{\dfrac{{{\left( 5-\sqrt{5} \right)}^{2}}}{\left( 5+\sqrt{5} \right)\left( 5-\sqrt{5} \right)}} \\\ \end{aligned}
⇒sin⁡π10=12(5−5)(5−5)(5+5)(5−5)\Rightarrow \sin \dfrac{\pi }{10}=\dfrac{1}{2}\sqrt{\dfrac{\left( 5-\sqrt{5} \right)\left( 5-\sqrt{5} \right)}{\left( 5+\sqrt{5} \right)\left( 5-\sqrt{5} \right)}}
We know that (a−b)2=a2−2ab+b2{{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}} and (a2−b2)=(a+b)(a−b)\left( {{a}^{2}}-{{b}^{2}} \right)=\left( a+b \right)\left( a-b \right) .
⇒sin⁡π10=12(5−5)2(5+5)(5−5) ⇒sin⁡π10=1225−105+525−5 ⇒sin⁡π10=1230−10520 ⇒sin⁡π10=123−52 \begin{aligned} & \Rightarrow \sin \dfrac{\pi }{10}=\dfrac{1}{2}\sqrt{\dfrac{{{\left( 5-\sqrt{5} \right)}^{2}}}{\left( 5+\sqrt{5} \right)\left( 5-\sqrt{5} \right)}} \\\ & \Rightarrow \sin \dfrac{\pi }{10}=\dfrac{1}{2}\sqrt{\dfrac{25-10\sqrt{5}+5}{25-5}} \\\ & \Rightarrow \sin \dfrac{\pi }{10}=\dfrac{1}{2}\sqrt{\dfrac{30-10\sqrt{5}}{20}} \\\ & \Rightarrow \sin \dfrac{\pi }{10}=\dfrac{1}{2}\sqrt{\dfrac{3-\sqrt{5}}{2}} \\\ \end{aligned}
Now, we can write z as
z=1212(5+5)+i123−52z=\dfrac{1}{2}\sqrt{\dfrac{1}{2}\left( 5+\sqrt{5} \right)}+i\dfrac{1}{2}\sqrt{\dfrac{3-\sqrt{5}}{2}}

When n=1n=1 ,
⇒θ=π10(1+4) ⇒θ=π10×5 ⇒θ=π2 \begin{aligned} & \Rightarrow \theta =\dfrac{\pi }{10}\left( 1+4 \right) \\\ & \Rightarrow \theta =\dfrac{\pi }{10}\times 5 \\\ & \Rightarrow \theta =\dfrac{\pi }{2} \\\ \end{aligned}
∴z=cos⁡π2+isin⁡π2 ⇒z=1+0 ⇒z=1 \begin{aligned} & \therefore z=\cos \dfrac{\pi }{2}+i\sin \dfrac{\pi }{2} \\\ & \Rightarrow z=1+0 \\\ & \Rightarrow z=1 \\\ \end{aligned}
When n=2n=2 ,
⇒θ=π10(1+4×2) ⇒θ=π10×(1+8) ⇒θ=9π10 \begin{aligned} & \Rightarrow \theta =\dfrac{\pi }{10}\left( 1+4\times 2 \right) \\\ & \Rightarrow \theta =\dfrac{\pi }{10}\times \left( 1+8 \right) \\\ & \Rightarrow \theta =\dfrac{9\pi }{10} \\\ \end{aligned}
We will get z as
z=cos⁡9π10+isin⁡9π10z=\cos \dfrac{9\pi }{10}+i\sin \dfrac{9\pi }{10}
We have to evaluate similar to the above steps. We will get
z=−1212(5+5)+i123−52z=-\dfrac{1}{2}\sqrt{\dfrac{1}{2}\left( 5+\sqrt{5} \right)}+i\dfrac{1}{2}\sqrt{\dfrac{3-\sqrt{5}}{2}}
When n=3n=3 ,
⇒θ=π10(1+4×3) ⇒θ=π10×(1+12) ⇒θ=13π10 \begin{aligned} & \Rightarrow \theta =\dfrac{\pi }{10}\left( 1+4\times 3 \right) \\\ & \Rightarrow \theta =\dfrac{\pi }{10}\times \left( 1+12 \right) \\\ & \Rightarrow \theta =\dfrac{13\pi }{10} \\\ \end{aligned}
We will get z as
z=cos⁡13π10+isin⁡13π10z=\cos \dfrac{13\pi }{10}+i\sin \dfrac{13\pi }{10}
We have to evaluate similar to the above steps. We will get
z=−1212(5+5)−i123−52z=-\dfrac{1}{2}\sqrt{\dfrac{1}{2}\left( 5+\sqrt{5} \right)}-i\dfrac{1}{2}\sqrt{\dfrac{3-\sqrt{5}}{2}}
When n=4n=4 ,
⇒θ=π10(1+4×4) ⇒θ=π10×(1+16) ⇒θ=17π10 \begin{aligned} & \Rightarrow \theta =\dfrac{\pi }{10}\left( 1+4\times 4 \right) \\\ & \Rightarrow \theta =\dfrac{\pi }{10}\times \left( 1+16 \right) \\\ & \Rightarrow \theta =\dfrac{17\pi }{10} \\\ \end{aligned}
We will get z as
z=cos⁡17π10+isin⁡17π10z=\cos \dfrac{17\pi }{10}+i\sin \dfrac{17\pi }{10}
We have to evaluate similar to the above steps. We will get
z=1212(5+5)−i123−52z=\dfrac{1}{2}\sqrt{\dfrac{1}{2}\left( 5+\sqrt{5} \right)}-i\dfrac{1}{2}\sqrt{\dfrac{3-\sqrt{5}}{2}}
This process repeats. Therefore, we will get six value of z as
z=0,i,±1212(5+5)+i123−52,z=±1212(5+5)−i123−52z=0,i,\pm \dfrac{1}{2}\sqrt{\dfrac{1}{2}\left( 5+\sqrt{5} \right)}+i\dfrac{1}{2}\sqrt{\dfrac{3-\sqrt{5}}{2}},z=\pm \dfrac{1}{2}\sqrt{\dfrac{1}{2}\left( 5+\sqrt{5} \right)}-i\dfrac{1}{2}\sqrt{\dfrac{3-\sqrt{5}}{2}}

Note: Students must know to simplify the exponents. They must know to find the values of trigonometric functions using the identities. Students must know that the complex number,z can be written as z=reiθz=r{{e}^{i\theta }} or z=x+iyz=x+iy.